A block of aluminum with a mass of 140 g is cooled from 98.4°C to 62.2°C with a release of 4817 J of heat. From these data, calc ulate the specific heat of aluminum?
1 answer:
Answer:
specific heat = 0.951 j/g·°C
Explanation:
Heat flow equation => q = m·c·ΔT
q = heat flow = 4817 joules
m = mass in grams = 140 grams Aluminum
c = specific heat = ?
ΔT = Temperature Change in °C = 98.4°C - 62.2°C = 36.2°C
q = m·c·ΔT => c = q/m·ΔT = 4817j/(140g)(36.2°C) = 0.951 j/g·°C
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